Problem: Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}-2x-8y &= 1 \\ -x+6y &= 3\end{align*}$
Answer: Begin by moving the $y$ -term in the second equation to the right side of the equation. $-x = -6y+3$ Divide both sides by $-1$ to isolate $x$ $x = {6y - 3}$ Substitute this expression for $x$ in the first equation. $-2({6y - 3}) - 8y = 1$ $-12y + 6 - 8y = 1$ Simplify by combining terms, then solve for $y$ $-20y + 6 = 1$ $-20y = -5$ $y = \dfrac{1}{4}$ Substitute $\dfrac{1}{4}$ for $y$ in the top equation. $-2x-8( \dfrac{1}{4}) = 1$ $-2x-2 = 1$ $-2x = 3$ $x = -\dfrac{3}{2}$ The solution is $\enspace x = -\dfrac{3}{2}, \enspace y = \dfrac{1}{4}$.